Question

The normal (perpendicular) line to the curve y=sqrt(8-x^2) at (-2,2) has slope
A. -2
B. 1/2
C. -1/2
D. 1
E. -1

I got A

Answer 1

y = √(8-x^2)
y' = -x/√(8-x^2)
y'(-2) = 2/2 = 1
so, the normal has slope -1